It is also confusing to see that the observed \(\eta _1(1855)\)  could be embedded into the compact tetraquark configuration [1]}, [2]}.
One reason for this is that the gluon of the hybrid can easily split into a quark pair, which forms a compact tetraquark state.
This imply that it is difficult to differentiate the hybrid mesons from the compact tetraquark states through the strong decay models.
The radiative decay may be a useful way to reveal their difference because the photon cannot interact directly with the gluon. Based on
the quark model, in which the decay of a hadron is assumed to proceed through a single quark transition, and the total decay width can be
obtained simply by summing the partial widths [3]}, we predict that if the \(\eta _1(1855)\)  is a compact tetraquark state,
the radiative decay width is two times larger than the results that are obtained by assuming \(\eta _1(1855)\)  as a hybrid meson\(^{[1]}\) .
[1]Here we consider the lowest order Feynman diagram. That means the \(\eta _1(1855)\)  only consists of valence quark if \(\eta _1(1855)\) 
is a compact tetraquark state or valence quark and valence gluon if \(\eta _1(1855)\)  is a hybrid meson.
Moreover, \(\eta _1(1855)\)  can also be considered as \(K\bar{K}_1(1400)\)  bound state [4]}, [5]}. Its molecular component
\(K\bar{K}_1(1400)\)  can be accurately estimated through the radiative decay width [4]}. A possible explanation for this may be
that \(\eta _1(1855)\)  interacts with the photons through its molecular component \(K\bar{K}_1(1400)\) , which is quite different from the quark
interacts directly with photons if \(\eta _1(1855)\)  is a compact tetraquark state or hybrid state.
